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Can this geometry probelm be solved?

  1. May 12, 2006 #1
    I had it on a math competition today and I couldn't think of anything. Express using variables.

    Any insights guys?

    http://img87.imageshack.us/img87/8684/math1wy.jpg [Broken]
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. May 12, 2006 #2


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    My first thought is tedius, but completely straightforward.

    Do you know the area of a sector of a circle? Can you find the area of a similar shape whose vertex is not at the center of the circle? If so, then you can find the area of this shaded region.
  4. May 12, 2006 #3
    ^ Sorry... I don't.

    I'm a high school senior if that matters.
  5. May 12, 2006 #4


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    Actually, it's a very simple calculation. You can just treat the required area as the sum of 4 equal segments (not sectors) of the circle plus the inscribed square.

    To find the area of one segment use the formula [tex]a = \frac{1}{2}r^2(\theta - \sin\theta)[/tex]. [tex]\theta[/tex] is labelled as angle PDQ in the attachment.

    To find [tex]\theta[/tex], observe that two of the quarter-circle arcs meet at a point corresponding to a vertical distance halfway along the side of the square (in order to preserve symmetry). An equilateral triangle PDC is formed by 3 lengths, one of them being the side of the circumscribing square (DC) (which is also the radius of the circle) and the others being the chords going from the corners of the square to the circle. Hence the angle subtended by the chord and the side of the square angle PDC is [tex]\frac{\pi}{3}[/tex]. [tex]\theta[/tex] can be calculated from this by subtracting [tex]\frac{\pi}{4}[/tex] and doubling, i.e. [tex]\theta = 2(\frac{\pi}{3} - \frac{\pi}{4}) = \frac{\pi}{6}[/tex].

    The rest of the calculation is easy. Just plug in the value of theta to find the area of one segment and multiply by four. To find the area of the inscribed square, use the side (given by [tex]2r\sin{(\frac{\pi}{12})}[/tex] and square that. Add it all up to give the required area.

    BTW, not to give away the exact answer, but I get mine as being slightly less than a third of the area of the big circumscribing square.

    Attached Files:

    Last edited: May 12, 2006
  6. May 13, 2006 #5
    now that the four arcs intersect relative to 4 centers of 4 edges
    You choose center of square is (0,0), then points are from -a to a.
    That problem can be solved by integrate y=mx^2+nx+p
    m,n,p are constants found by picking up 3 special points
    finally multiply 4 to find the result
  7. May 13, 2006 #6


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    Isn't that overkill for this problem?
  8. May 13, 2006 #7


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    I got 1 + pi/3 - sqrt(3).

    Thats assuming its a unit square of course (othewise just multiply by the area of the square).

    I used :
    Area = 2 times Area "()" plus 4 times Area "/\" minus 1 times Area of outer square.

    Sorry about the cryptic descriptions of the regions but I'm too lazy to draw a proper diagram. Area "()" refers to the region that looks a bit like that and "/\" refers to the little inverted V shaped region with base BC.
  9. May 13, 2006 #8


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    This is correct, I found the same answer through a geometrical approach.

    Check using calculus:

    8\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\int\limits_{\frac{1}{{2\sin t}}}^1 {rdr} dt} = 8\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\left[ {\frac{{r^2 }}{2}} \right]_{\frac{1}{{2\sin t}}}^1 dt} = 8\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\frac{1}{2} - \frac{1}{{8\sin ^2 t}}dt} = 8\left[ {\frac{t}{2} + \frac{{\cot t}}{8}} \right]_{\frac{\pi }{6}}^{\frac{\pi }{4}} \\ \\
    = 8\left( {\left( {\frac{\pi }{8} + \frac{1}{8}} \right) - \left( {\frac{\pi }{{12}} + \frac{{\sqrt 3 }}{8}} \right)} \right) = 8 \cdot \frac{{\pi - 3\sqrt 3 + 3}}{{24}} = \frac{\pi }{3} - \sqrt 3 + 1 \approx 0.315 \\

    I integrated over the http://www.td-hosting.com/wisfaq/cirkels.gif" [Broken], multiplied by 8.
    Last edited by a moderator: May 2, 2017
  10. May 13, 2006 #9


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    Perhaps the 3D-equivalent is a bit more challenging.
    A unit cube with a unit sphere on each vertex. Find the common volume.

    Or would it even be possible to generalize this to n-cubes with n-spheres, creating a common n-volume? This is over my head :redface:
  11. May 14, 2006 #10
    Thank You very much!

    Can it be solved without Calculus?
  12. May 14, 2006 #11


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    Yes! For example, see my post.:smile:
  13. May 16, 2006 #12
    Just keep in mind the symmetry of this problem...it should help significantly.
  14. May 18, 2006 #13
    Well, there's something i'm not getting on this problem. basicly i found the intersection between each 2 quarters of circle:

    A() = 2A0/4 - A[] = A0/2 - A []

    then i found the common area between the 4 quarters:

    Ac = A0 - A[]

    Now, from what i think:

    Ac = 2*A() + 2*A

    However i'm getting A= (r^2)/2 which can't be true.

    Where am i wrong?
  15. May 22, 2006 #14
    You guys are right...

    The answer was given to me today
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